A ball thrown up vertically returns to the thrower after 6s. Find… - Vidyadip

Question

A ball thrown up vertically returns to the thrower after $6s$. Find its position after $4s$.

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Answer

Ball attains the maximum height after 3s. After attaining this height, it will start falling downwards.
In this case,
Initial velocity, $u = 0$
Position of the ball after 4s of the throw is given by the distance travelled by it
during its downward journey in 4s - 3s = 1s.
Equation of motion, $s = ut + 1/2 gt^2$ will give,
$s = 0 \times t + 1/2 \times 9.8 \times 1^2 = 4.9m$
Total height $= 44.1m$
This means that the ball is $39.2m (44.1m - 4.9m)$ above the ground after $4$ seconds.

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